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Standard XII

Chemistry

Stoichiometric Calculations

A 10 g mixt...

Question

A $10$ g mixture of $C u_{2} S$ and $C u S$ was treated with $200$ mL of $0.75 M M n O_{4}^{-}$ in acid solution producing $S O_{2}$ , $C u^{2 +}$ and $M n^{2 +}$ . The $S O_{2}$ was boiled off and the excess of $M n O_{4}^{-}$ was titrated with $175$ mL of $1 M F e^{2 +}$ solution. Calculate the percentage of $C u S$ in the original mixture (to nearest integer).

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Solution

Meq. of $M n O_{4}^{-}$ added $= 200 \times 0.75 \times 5 = 750$
$M n^{7 +} + 5 e ⟶ M n^{2 +}$ $(∴ N = M \times 5)$
Meq. of $M n O_{4}^{-}$ left unused = Meq. of $F e^{2 +}$ used $= 175 \times 1 \times 1 = 175$
$F e^{2 +} ⟶ F e^{3 +} + e$ $(∴ N = M \times 1)$
Now, meq. of $M n O_{4}^{-}$ used $= 750 - 175 = 575$
$M n O_{4}^{-}$ is used for $C u_{2} S$ and $C u S$ .
For $C u_{2} S$ :
$(C u^{+})_{2} ⟶ 2 C u^{2 +} + 2 e$
$S^{2 -} ⟶ S^{4 +} + 6 e$
$C u_{2} S ⟶ 2 C u^{2 +} + S^{4 +} + 8 e$
For $C u S$ :
$S^{2 -} ⟶ S^{4 +} + 6 e$
Let masses of $C u_{2} S$ and $C u S$ be $a$ and $b$ respectively.
$∴ a + b = 10$ $. . . (i)$
$∴$ Meq. of $M n O_{4}$ used $=$ Meq. of $C u_{2} S +$ Meq. of $C u S$
$575 = \frac{a}{(\frac{159.2}{8})} \times 1000 + \frac{b}{(\frac{95.6}{6})} \times 1000$ $. . . (i i)$
$S o l v i n g E q s .$ (i) $a n d$ (ii) $, w e g e t$ a=4.206 $g a n d$ b=5.794 $g$ \therefore\:\% $o f$ CuS $i n t h e m i x t u r e$ =\displaystyle\frac{5.794}{10}\times100=57.94\%$

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A 10 g mixture of Cu2S and CuS was treated with 200 mL of 0.75MMnO−4 in acid solution producing SO2, Cu2+ and Mn2+. The SO2 was boiled off and the excess of MnO−4 was titrated with 175 mL of 1MFe2+ solution. Calculate the percentage of CuS in the original mixture (to nearest integer).